Calculate the solubility in moles per liter of Hg2Cl2,
Ksp = 1.10 x 10-18 ,
Hg2Cl2(aq) ---> Hg22+ + 2Cl-
2007-11-01 07:34:57 · 3 answers · asked by mimi 1 in Science & Mathematics ➔ Chemistry
So you know that Ksp = [Hg22+][Cl-]^2.
So, let [Hg22+] = x. The [Cl-] will be 2x since you form two of these for every mercury ion. Substitute those into the expression for Ksp:
Ksp = 1.10 X 10^-18 = (x) X (2x)^2 = 4x^3.
Just solve for x and that will be your solubility in mol/L.
2007-11-01 07:39:45 · answer #1 · answered by hcbiochem 7 · 0⤊ 0⤋
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2016-05-17 03:18:01 · answer #2 · answered by Lori 4 · 0⤊ 0⤋
Ksp = [Hg2++][Cl-]^2
Let the molar solubility be x. Then [Hg2++] = x and [Cl-] = 2x
(x)(2x)^2 = 1.10 x 10^-18
4x^3 = 1.10 x 10^-18
x^3 = 4400 x 10^-21
x = 1466 x 10^-7 = 1.47 x 10^-4 moles/L
2007-11-01 07:41:30 · answer #3 · answered by steve_geo1 7 · 0⤊ 0⤋